I am working to an index that resumes many of the ideas showed in the precedents post
I = m log(d t) : log((n+e) c)
m=multeplicity: 2 for normal hits, 3 for triples, 4 for quadruples...
d=distance in km in d>=1 ; d=1 if the distance is 0 km
t= time in days
n= number of hit involving the users
c= number of hit involving the locations
e= little number because log(1) = 0
in my calculations , e is for the moment 0,1 , but we will see later about that number
log = decimal logarithm (because is more easy do calculations, we can change the base)
About the numerator. Usually the kilometres are something about 10, 100, 500, 1000 if you are very lucky 2000 or 3000, but hit of 15 000 km are extremely unusual. Usually the days are 10, 100, 500, more than 1000 they became unusual. So the product is something between 1 and about 3 millions: with the log it becomes something between 1 and 6.
About the denominator: it's quite difficult to have "usual values" here, but in this formula, when c is locked, the increase of 10 times of the number of hits between two users may be balanced by the increase of 10 times of the distance or days, so for example
Wien-Wien 7 days between two little users = Wien-Wien 70 days between two users with 10 times hits togheter
The problem is the zero.
On the numerator: 1 or 0 km are approximatively the same thing, so we can replace the 0 with a 1; 0 days it is not possible (automatically moderated), so no problems.
On the denominator: here 1 or 2 hits between users or cities ARE different, and we have to forbid that the product will be 1 because of the logarithm, so I placed temporaneally a little number, e, for correct that. In my first calculation this number was 0,1 , but I saw that if e = 0,96478 , a hit with 10 days and 10 km, n=1, c=1, has I =100, and that is nice
I tried this index with repetite hits in the time, e=0,1
1) User A inserts some notes from a place. Those notes are found by user B at 10 km of distance, the first the day 1, the second the day 10, the third the day 20, the fourth the day 30, the fifth the day 40
I = 48,32 , then 6,418, then 4,752, then 4,078, then 3,700
2) User A inserts some notes from a place. Those notes are found by user B always on days 1,10, 20... but in different locations (user B is rotating around A)
I=48,32, then 12,41, then 9,366 , then 8,084, then 7,355
3)User A and user B are moving around Europe, but user B found many bills of A, (d t) is always 100, but locations are always virgins
I= 96,64 , then 12,41, then 8,140 , then 6,528, then 5,655
Now I have to test this algorithm with real hits, but I have got the problem of finding n and c